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A210640 a(n) = least integer m > 1 such that 2S_k^2 (k=1,...,n) are pairwise distinct modulo m, where S_k is the sum of the first k primes.

%I A210640 #15 Apr 18 2013 02:53:45
%S A210640 2,4,9,13,17,28,37,37,37,37,37,61,61,61,151,151,151,151,151,151,151,
%T A210640 227,227,227,227,227,307,307,307,337,433,433,433,433,433,433,433,433,
%U A210640 433,433,433,509,509,509,509,509,643,727,727,761,761,761,971,971,971
%N A210640 a(n) = least integer m > 1 such that 2S_k^2 (k=1,...,n) are pairwise distinct modulo m, where S_k is the sum of the first k primes.
%C A210640 If a(n) is an odd prime p_k then a(n)|2(S_k^2-S_{k-1}^2) and hence k>n. Zhi-Wei Sun conjectured that a(n) is a prime smaller than n^2 unless n divides 6.
%H A210640 Zhi-Wei Sun, <a href="/A210640/b210640.txt">Table of n, a(n) for n = 1..5258</a>
%H A210640 Zhi-Wei Sun, <a href="http://dx.doi.org/10.1016/j.jnt.2013.02.003">On functions taking only prime values</a>, J. Number Theory 133(2013), no.8, 2794-2812.
%H A210640 Zhi-Wei Sun, <a href="https://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;15b3d194.1203">On sums of consecutive primes</a>, a message to Number Theory List, March 23, 2012.
%e A210640 We have a(3)=9, because 2*2^2=8, 2*(2+3)^2=50, 2(2+3+5)^2=200 are pairwise distinct modulo m=9 but not pairwise distinct modulo any of 2, 3, 4, 5, 6, 7, 8.
%t A210640 s[n_] := s[n] = Sum[Prime[k], {k,1,n}]; f[n_] := f[n] = 2*s[n]^2; R[n_,m_] := Union[Table[Mod[f[k],m], {k,1,n}]]; Do[Do[If[Length[R[n,m]] == n, Print[n," ",m]; Goto[aa]], {m,2,Max[2,n^2]}]; Print[n]; Label[aa]; Continue, {n,1,5000}]
%Y A210640 Cf. A000040, A210393, A210394, A210186, A210144, A208494, A208643, A207982.
%K A210640 nonn
%O A210640 1,1
%A A210640 _Zhi-Wei Sun_, Mar 26 2012